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If log⁡75=p\log_7{5}=p and log⁡102=q\log_{10}{2} = q, solve log⁡72\log_7{2} in terms of pp and qq

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Full solution

Q. If log⁡75=p\log_7{5}=p and log⁡102=q\log_{10}{2} = q, solve log⁡72\log_7{2} in terms of pp and qq
  1. Understand Given Information: Understand the given information and what is being asked.\newlineWe are given that log⁡75=p\log_7{5}=p and log⁡102=q\log_{10}{2} = q. We need to express log⁡72\log_7{2} in terms of pp and qq.
  2. Express Exponential Equations: Express log⁡75=p\log_7 5 = p as an exponential equation.\newline7p=57^p = 5
  3. Relate Logarithms: Express log⁡102=q\log_{10}2=q as an exponential equation.\newline10q=210^q = 2
  4. Break Down Expression: Express log⁡72\log 7^2 in terms of pp. We need to find a way to relate log⁡72\log 7^2 to log⁡75\log 7^5. We can do this by recognizing that 22 is a factor of 55, specifically 5=2×(5/2)5 = 2 \times (5/2). So we can write: log⁡72=log⁡7(5×(2/5))\log 7^2 = \log 7(5 \times (2/5))
  5. Simplify Using Information: Use the properties of logarithms to break down log⁡7(5⋅(2/5))\log_7(5 \cdot (2/5)).log⁡72=log⁡7(5)+log⁡7(2/5)\log_7^2 = \log_7(5) + \log_7(2/5)
  6. Substitute Known Values: Simplify the expression using the given information.\newlineWe know that 7p=57^p = 5, so log⁡7(5)=p\log_7(5) = p. Now we need to express log⁡7(25)\log_7\left(\frac{2}{5}\right) in terms of pp and qq.\newlinelog⁡72=p+log⁡7(25)\log_7^2 = p + \log_7\left(\frac{2}{5}\right)
  7. Change Base: Break down log⁡7(25)\log_7\left(\frac{2}{5}\right) further using the properties of logarithms.\newlinelog⁡7(25)=log⁡7(2)−log⁡7(5)\log_7\left(\frac{2}{5}\right) = \log_7(2) - \log_7(5)
  8. Simplify Further: Substitute the known values into the equation.\newlineWe know that log⁡7(5)=p\log_7(5) = p, so we can substitute that in. We also know that 10q=210^q = 2, which means log⁡10(2)=q\log_{10}(2) = q. We need to change the base from 1010 to 77 for log⁡7(2)\log_7(2).\newlinelog⁡72=p+(log⁡7(2)−p)\log_7^2 = p + (\log_7(2) - p)
  9. Substitute Back: Change the base of log⁡10(2)\log_{10}(2) to log⁡7(2)\log_{7}(2) using the change of base formula.log⁡7(2)=log⁡10(2)log⁡10(7)\log_{7}(2) = \frac{\log_{10}(2)}{\log_{10}(7)}Since log⁡10(2)=q\log_{10}(2) = q, we can substitute qq for log⁡10(2)\log_{10}(2).log⁡7(2)=qlog⁡10(7)\log_{7}(2) = \frac{q}{\log_{10}(7)}
  10. Final Simplification: Substitute log⁡7(2)\log_7(2) back into the equation for log⁡72\log_7^2.
    log⁡72=p+(qlog⁡10(7)−p)\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right)
  11. Final Simplification: Substitute log⁡7(2)\log_7(2) back into the equation for log⁡72\log_7^2.
    log⁡72=p+(qlog⁡10(7)−p)\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right) Simplify the expression.
    Since p−p=0p - p = 0, we can remove them from the equation.
    log⁡72=qlog⁡10(7)\log_7^2 = \frac{q}{\log_{10}(7)}
  12. Final Simplification: Substitute log⁡7(2)\log_7(2) back into the equation for log⁡72\log_7^2.
    log⁡72=p+(qlog⁡10(7)−p)\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right) Simplify the expression.
    Since p−p=0p - p = 0, we can remove them from the equation.
    log⁡72=qlog⁡10(7)\log_7^2 = \frac{q}{\log_{10}(7)} Recognize that log⁡10(7)\log_{10}(7) is simply the reciprocal of log⁡7(10)\log_7(10).
    log⁡72=q(1/log⁡7(10))\log_7^2 = \frac{q}{(1 / \log_7(10))}
  13. Final Simplification: Substitute log⁡7(2)\log_7(2) back into the equation for log⁡72\log_7^2.
    log⁡72=p+(qlog⁡10(7)−p)\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right) Simplify the expression.
    Since p−p=0p - p = 0, we can remove them from the equation.
    log⁡72=qlog⁡10(7)\log_7^2 = \frac{q}{\log_{10}(7)} Recognize that log⁡10(7)\log_{10}(7) is simply the reciprocal of log⁡7(10)\log_7(10).
    log⁡72=q(1/log⁡7(10))\log_7^2 = \frac{q}{(1 / \log_7(10))} Simplify the expression by multiplying by the reciprocal.
    log⁡72=q⋅log⁡7(10)\log_7^2 = q \cdot \log_7(10)
  14. Final Simplification: Substitute log⁡7(2)\log_7(2) back into the equation for log⁡72\log_7^2.
    log⁡72=p+(qlog⁡10(7)−p)\log_7^2 = p + \left(\frac{q}{\log_{10}(7)} - p\right) Simplify the expression.
    Since p−p=0p - p = 0, we can remove them from the equation.
    log⁡72=qlog⁡10(7)\log_7^2 = \frac{q}{\log_{10}(7)} Recognize that log⁡10(7)\log_{10}(7) is simply the reciprocal of log⁡7(10)\log_7(10).
    log⁡72=q1/log⁡7(10)\log_7^2 = \frac{q}{1 / \log_7(10)} Simplify the expression by multiplying by the reciprocal.
    log⁡72=q⋅log⁡7(10)\log_7^2 = q \cdot \log_7(10) Recognize that log⁡7(10)\log_7(10) is a constant and cannot be simplified further in terms of log⁡72\log_7^200 and log⁡72\log_7^211.
    Therefore, the final expression for log⁡72\log_7^2 in terms of log⁡72\log_7^200 and log⁡72\log_7^211 is:
    log⁡72=q⋅log⁡7(10)\log_7^2 = q \cdot \log_7(10)

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